# Estmated time required for my following program

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taher azim on 23 Sep 2016
Commented: Steven Lord on 23 Sep 2016
Can anyone help me out with their past experience that how much time this below program will take approximately containing 6 for loops with each performing approx. 15 itterations.Matlab is utilizing 985 Mb of ram.
function o=optimum() m=[1,1.25,1.5,2,2.5,3,4,5,6,8]; %normal module from psg q=1; l=zeros(4500000,16); j1=input('\n enter j'); Pin=input('\ninput power'); ip=input('\n reduction'); N1=input('\n input speed'); sp1=700; sg1=300; Ep=205000; Eg=205000; BHN=300;
for w=9:1:23 %helical angle for gearpair 1
for g=2.1:0.3:8.1 %gear ratio for first gearpair
a=(2*(cos(w*pi/180)^3))/(sin(pi/9)^2); %condition for min no teeth on 1st pinion
t=ceil(a);
for t1=t:1:t+9 %no of teeth on first gear
t2=round(g*t1); %no of teeth on second gear
gmdf=t2/t1; %modified gear ratio of 1st pair
mt=m(1,1)/cos(w*pi/180);%transverse module of 1st pair
d1pcd=mt*t1;%pcd of 1st gear
d2pcd=mt*t2;%pcd of second gear
for i=1:1:10
for w1=9:1:23 %helical angle for gearpair 2
g2=ip/gmdf; %ip is the input gear reduction by user
c=(2*(cos(w*pi/180)^3))/(sin(pi/9)^2); %condition for min no teeth on 2nd pinion
z=ceil(c);
for t3=z:1:z+9 %no of teeth on 3rd gear
t4=round(g2*t3); %no of teeth on 4th gear
gmdf1=t4/t3; %modified gear ratio of 2nd pair t3; %
mt1=m(1,i)/cos(w1*pi/180); %transverse module of second pair
d4pcd=mt1*t4; %pcd of 4th gear
d3pcd=mt1*t3; %pcd of 3rd gear
v1=(pi*d1pcd*N1)/(60*1000); %N1 is the input speed by the user
v4=v1/(gmdf*gmdf1); %pitch line velocity of second gear pair
b1=ceil((1.15*pi*m(1,1))/sin(w*pi/180)); %face width first gear pair
if b1>15*m(1,1) && b1<9*m(1,1)
b1=round(10.5*m(1,1));
else
b1=ceil((1.15*pi*m(1,1))/sin(w*pi/180));
end
b2=ceil((1.15*pi*m(1,i))/sin(w1*pi/180)); %face width of second gear pair
if b2>15*m(1,i) && b2<9*m(1,i)
b2=round(10.5*m(1,i));
else
b2=ceil((1.15*pi*m(1,i))/sin(w1*pi/180));
end
fb11=sp1*b1*m(1,1)*(0.484-(2.87*cos(w)^3)/(t1)); % sp1 is bending stress
fb12=sg1*b1*m(1,1)*(0.484-(2.87*cos(w)^3)/(t2)); % fb is bending stress
if fb11>fb12
fb1=fb12;
else
fb1=fb11;
end
fb13=sp1*b2*m(1,i)*(0.484-(2.87*cos(w1)^3)/(t3)); %sp=ultimate tensile strength/3(of pinion)
fb14=sg1*b2*m(1,i)*(0.484-(2.87*cos(w1)^3)/(t4)); %sg=ultimate tensile strength/3(of gear)
if fb13>fb14;
fb2=fb14;
else
fb2=fb13;
end
q1=(2*t2)/(t1+t2);
q2=(2*t4)/(t3+t4);
k=(0.18)*(BHN/100)^2;
fw1=(d1pcd*b1*q1*k)/(cos(w*pi)/180); %fw is surface wear force
fw2=(d3pcd*b2*q2*k)/(cos(w1*pi)/180);
e1=(16+0.63*(m(1,i)+0.25*(d1pcd)^(1/2))+0.63*(m(1,i)+0.25*(d2pcd)^(1/2)))*0.001; %pitch error on meshing teeth
c1=(0.111*e1)*((Ep*Eg)/(Ep+Eg)); %Ep=modulus of elasticity of pinion and Eg=modulus of elasticity of gear
Ft1=1.25*1.5*(Pin/v1); %first factor of f effective and Pin=input power by user
Fd1=(21*v1*cos(w*pi/180)*(b1*c1*(cos(w*pi/180))^2+Ft1))/(21*v1+(b1*c1*(cos(w*pi/180))^2+Ft1)^(1/2));
Feff1=Ft1+Fd1; % effective force of first gear pair
e2=(16+0.63*(m(1,i)+0.25*(d3pcd)^(1/2))+0.63*(m(1,i)+0.25*(d4pcd)^(1/2)))*0.001; %pitch error on meshing teeth
c2=(0.111*e2)*((Ep*Eg)/(Ep+Eg));
Ft2=1.25*1.5*(Pin/v4);
Fd2=(21*v4*cos(w1*pi/180)*(b2*c2*(cos(w1*pi/180))^2+Ft2))/(21*v4+(b2*c2*(cos(w1*pi/180))^2+Ft2)^(1/2));
Feff2=Ft2+Fd2; %effective force on second gearpair
cd1=(d1pcd+d2pcd)/2;
cd2=(d3pcd+d4pcd)/2;
cd=max(cd1,cd2);
m1=m(1,i);
if fb1>Feff1 && fb2>Feff2 && fw1>Feff1 && fw2>Feff2
f.o.s1=(fb1/Feff1); %factor of safety for first pair (bending)
f.o.s2=(fb2/Feff2); %factor of safety for second pair (bending)
f.o.s3=(fw1/Feff3); %factor of safety for first pair (surface)
f.o.s4=(fw2/Feff4); %factor of safety for second pair (surface)
x=[f.o.s1,f.o.s2,f.o.s3,f.o.s4];
z=min(x);
if j1==1
u=1;
v=0;
l(q,:)=[m(1,1) w gmdf t1 t2 m1 w1 gmdf1 t3 t4 cd1 cd2 cd u v z];
q=q+1;
elseif j2==1
if z>r %r=fos given by user
u=1; %u=logical array for force
v=1; %v=logical array for fos
l(q,:)=[m(1,1) w gmdf t1 t2 m1 w1 gmdf1 t3 t4 cd1 cd2 cd u v z];
q=q+1;
else
u=1;
v=0;
l(q,:)=[m(1,1) w gmdf t1 t2 m1 w1 gmdf1 t3 t4 cd1 cd2 cd u v z];
q=q+1;
end
end
else
u=0;
v=0;
z=0;
l(q,:)=[m(1,1) w gmdf t1 t2 m1 w1 gmdf1 t3 t4 cd1 cd2 cd u v z];
q=q+1;
end
end
end
end
end
end
end
if j1==1
t=find(l(:,14)==1);
n=size(t);
for i=1:1:n(1,1)
t(i,1)=l(t(i,1),13);
h=min(t);
end
elseif j2==1
t=find(l(:,15)==1);
n=size(t);
for i=1:1:n(1,1)
t(i,1)=l(t(i,1),13);
h=min(t);
end
end
o=find(l(:,13)==h);
fprintf('\n %f',o(1,1));
end

#### 1 Comment

I doubt anyone has past experience of running that precise code so it is impossible to answer. The number of nested for loops does not provide any indication as to how long it will take. I could write code that contains 10 nested for loops that completes in under a second or code with 1 for loop that carries on infinitely.

Edited: Adam on 23 Sep 2016
If you are using R2016a (or maybe R2015b, I can't remember when it was introduced) there is now the very useful pause feature when running code so that you don't have to just sit there and wonder if your code will ever end, you can pause it and land in the code at the current point to see how far it has got.
On the same note, you could use a progress bar (e.g. waitbar, though I prefer to use one of the many alternatives in the file exchange) around one of your for loops. This will give you an indication of how fast your code is going although this again is an estimate that depends very much on the nature of your code.
I like to use this one personally
as it comes with an estimated completion time, but you must be aware that the estimate is based purely on what proportion it has done so far - e.g. if you update it in a for loop from 1 to 10 then when it gets to the end of iteration 5 it's estimate of time remaining will be roughly equal to the time it has spent so far. It does not know if, for example, each iteration of your code gets slower and slower, in which case its estimate of time remaining will keep rising even as it progresses.
If you think your algorithm is fairly linear in its progress then you can run it with this progress bar for a certain period and then get an estimate of how long the full program would take. Obviously the longer you let it run for the more accurate an estimate it would be.

#### 1 Comment

Steven Lord on 23 Sep 2016
That feature was introduced in release R2016a.

Luuk van Oosten on 23 Sep 2016
Dear Taher,
How long such a programm will take to run depends on many things (both you hardware & your software). But... you can simply test how long it will take by using the stopwatch
tic
your_program
toc

taher azim on 23 Sep 2016
Actually I want an estimated time beforehand so that i can complete my project in required time.I had run the program yesterday and it had taken all night and still with no output. So was asking someone has encountered the same problem and if yes how much time it has taken for them to get their output. My hardware specifiactions- RAM-4 GB intel i5 processor 4th generation.
Luuk van Oosten on 23 Sep 2016
Dear Taher,
This is [in general] impossible, as Adam has already said. You could time individual parts of your loops and make an estimated guess.
One thing I found was this, which might assist you in estimating the run time.
Even with the [very rough] knowledge of your hardware, there are many factors in play which influence run time. The best thing to estimate run time is.... experience. If I were you, I would just run the program with tic-toc. Start on Friday afternoon, go home, have a beer, and see Monday morning what has happened.
taher azim on 23 Sep 2016
Thank u for your suggestion just ordering a beer now..